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Automorphism Group of the Commutator Subgroup of the Braid Group ANNALES DE LA FACULTÉ DES SCIENCES Mathématiques STEPAN YU.OREVKOV Automorphism group of the commutator subgroup of the braid group Tome XXVI, no 5 (2017), p. 1137-1161. <http://afst.cedram.org/item?id=AFST_2017_6_26_5_1137_0> © Université Paul Sabatier, Toulouse, 2017, tous droits réservés. L’accès aux articles de la revue « Annales de la faculté des sci- ences de Toulouse Mathématiques » (http://afst.cedram.org/), implique l’accord avec les conditions générales d’utilisation (http://afst.cedram. org/legal/). Toute reproduction en tout ou partie de cet article sous quelque forme que ce soit pour tout usage autre que l’utilisation à fin strictement personnelle du copiste est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright. cedram Article mis en ligne dans le cadre du Centre de diffusion des revues académiques de mathématiques http://www.cedram.org/ Annales de la faculté des sciences de Toulouse Volume XXVI, no 5, 2017 pp. 1137-1161 Automorphism group of the commutator subgroup of the braid group (∗) Stepan Yu. Orevkov (1) 0 ABSTRACT.— Let Bn be the commutator subgroup of the braid 0 group Bn. We prove that Aut(Bn) = Aut(Bn) for n > 4. This answers a question asked by Vladimir Lin. 0 RÉSUMÉ.— Soit Bn le groupe dérivé du groupe de tresses Bn. On 0 montre que Aut(Bn) = Aut(Bn) pour n > 4, ce qui répond à une question posée par Vladimir Lin. 1. Introduction Let Bn be the braid group with n strings. It is generated by σ1, . , σn−1 (called standard or Artin generators) subject to the relations σiσj = σjσi for |i − j| > 1 ; σiσjσi = σjσiσj for |i − j| = 1. 0 Let Bn be the commutator subgroup of Bn. Vladimir Lin [16] posed a prob- 0 lem to compute the group of automorphisms of Bn. In this paper we solve this problem. Theorem 1.1. — If n > 4, then the restriction mapping Aut(Bn) → 0 Aut(Bn) is an isomorphism. ∼ Dyer and Grossman [5] proved that Out(Bn) = Z2 for any n. The only nontrivial element of Out(Bn) corresponds to the automorphism Λ −1 defined by σi 7→ σi for all i = 1, . , n − 1. The center of Bn is gener- 2 Qn−1 Qn−i ated by ∆ where ∆ = ∆n = i=1 j=1 σj is Garside’s half-twist. Thus (*) Reçu le 11 novembre 2015, accepté le 5 juillet 2016. (1) Institut de Mathématiques de Toulouse, Université Paul Sabatier, 118 route de Narbonne, 31062 Toulouse, France — Steklov Mathematical Institute, 8 Gubkina St., 119991 Moscow, Russia — [email protected] I am grateful to the referee for many very useful remarks. Article proposé par Jean-Pierre Otal. – 1137 – Stepan Yu. Orevkov ∼ 2 Aut(Bn) = (Bn/h∆ i) o Z2. For an element g of a group, we denote the inner automorphism x 7→ gxg−1 by g˜. 0 Corollary 1.2. — If n > 4, then Out(Bn) is isomorphic to the dihedral group Dn(n−1) = Zn(n−1) o Z2. It is generated by Λ and σ˜1 subject to the 2 n(n−1) defining relations Λ =σ ˜1 = Λ˜σ1Λ˜σ1 = id. 0 For n = 3, the situation is different. It is proven in [11] that B3 is a free −1 −1 group of rank two generated by u = σ2σ1 and t = σ1 σ2 (in fact, the free 0 −1 base of B3 considered in [11] is u, v with v = t u). So, its automorphism group is well-known (see [18, §3.5, Theorem N4]). In particular (see [18, Corollary N4]), there is an exact sequence 0 ι 0 π 1 / B3 / Aut(B3) / GL(2, Z) / 1 0 where ι(x) =x ˜ and π takes each automorphism of B3 to the induced auto- 0 2 morphism of the abelianization of B3 (which we identify with Z by choos- −1 −1 ing the images of u and t as a base). We have σ˜1(u) = t u, σ˜2(u) = ut , −1 −1 σ˜1(t) =σ ˜2(t) = u, Λ(u) = t , Λ(t) = u whence 1 1 0 −1 π(˜σ ) = π(˜σ ) = , π(Λ) = . 1 2 −1 0 −1 0 0 ∼ Thus, again (as in the case n > 4) the image of Aut(B3) in Out(B3) = 0 GL(2, Z) is isomorphic to D6 but this time it is not the whole group Out(B3). Let Sn be the symmetric group and An its alternating subgroup. Let µ = µn : Bn → Sn be the homomorphism which takes σi to the transposition 0 0 (i, i + 1) and let µ be the restriction of µ to Bn. Then Pn = ker µn is the 0 0 0 group of pure braids. Let Jn = Bn ∩ Pn = ker µ . Note that the image of µ is An. The following diagram commutes where the rows are exact sequences and all the unlabeled arrows (except “→ 1”) are inclusions: 0 0 µ 1 / Jn / Bn / An / 1 (1.1) µ 1 / Pn / Bn / Sn / 1 Recall that a subgroup of a group G is called characteristic if it is invariant under each automorphism of G. Lin proved in [15, Theorem D] that Jn is a 0 characteristic subgroup of Bn for n > 5 (note that this fact is used in our proof of Theorem 1.1 for n > 5). By Theorem 1.1, this result extends to the case n = 4. – 1138 – Automorphism group of the commutator subgroup of the braid group 0 Corollary 1.3. — J4 is a characteristic subgroup of B4. 0 Note that J3 is not a characteristic subgroup of B3. Indeed, let ϕ ∈ 0 −1 Aut(B3) be defined by u 7→ u, t 7→ ut. Then ut ∈ J3 whereas ϕ (ut) = t 6∈ J3. 2. Preliminaries Let e : Bn → Z be the homomorphism defined by e(σi) = 1 for all 0 i = 1, . , n. Then we have Bn = ker e. 2.1. Groups For a group G, we denote its unit element by 1, the center by Z(G), the commutator subgroup by G0, the second commutator subgroup (G0)0 by G00, and the abelianization G/G0 by Gab. We denote x−1yx by yx (thus x˜(yx) = y) and we denote the commutator xyx−1y−1 by [x, y]. For g ∈ G, we denote the centralizer of g in G by Z(g, G). If H is a subgroup of G, then, evidently, Z(g, H) = Z(g, G) ∩ H. Lemma 2.1. — Let G be a group generated by a set A. Assume that there exists a homomorphism e : G → Z such that e(A) = {1}. Let e¯ be the induced ab homomorphism G → Z. Let Γ be the graph such that the set of vertices is A and two vertices a and b are connected by an edge when [a, b] = 1. If the graph Γ is connected, then (ker e)ab =∼ kere ¯. Proof. — Let K = ker e. Let us show that G0 ⊂ K0. Since K0 is normal in G, and G0 is the normal closure of the subgroup generated by [a, b], a, b ∈ A, it is enough to show that [a, b] ∈ K0 for any a, b ∈ A. We define a relation ∼ on A by setting a ∼ b if [a, b] ∈ K0. Since Γ is connected, it remains to note that this relation is transitive. Indeed, if a ∼ b ∼ c, then [a, c] = [a, b][bab−2, bcb−2][b, c] ∈ K0. Thus G0 ⊂ K0 whence G0 = K0 and we obtain Kab = K/K0 = K/G0 = kere ¯. 00 0 Remark 2.2. — The fact that Bn = Bn for n > 5 proven by Gorin and Lin [11] (see also [15, Remark 1.10]) is an immediate corollary of Lemma 2.1. n−1 Indeed, if we set G = Bn and A = {σi}i=1 , then Γ is connected whence 0 00 ab ∼ 00 0 Bn/Bn = (ker e) = kere ¯ = {1}. In the same way we obtain G = G when G is an Artin group of type Dn (n > 5), E6, E7, E8, F4, or H4. – 1139 – Stepan Yu. Orevkov 2.2. Pure braids 2 Recall that Pn is generated by the braids σij, 1 6 i < j 6 n, where −1 −1 σij = σji = σj−1 . σi+1σiσi+1 . σj−1. For a pure braid X, let us denote the linking number of the i-th and j-th strings by lkij(X). If X is presented by a diagram with under- and over-crossings, then lkij(X) is the half-sum of the signs of those crossings where the i-th and j-th strings cross. Let Aij 2 ab be the image of σij in Pn . We have, evidently, γ lkγ(i),γ(j)(X) = lki,j(X ), for any X ∈ Pn, γ ∈ Bn (2.1) (here γ(i) = µ(γ)(i) which is coherent with the interpretation of Bn as a mapping class group; see Section 4.1). ab It is well known that Pn is freely generated by {Aij}16i<j6n. This fact is usually derived from Artin’s presentation of Pn (see [1, Theorem 1.18]) but it also admits a very simple self-contained proof based on the linking numbers. Namely, let L be the free abelian group with a free base {aij}1 i<j n. Then P 6 6 it is immediate to check that the mapping Pn → L, X 7→ i<j lki,j(X)aij ab is a homomorphism and that the induced homomorphism Pn → L is the ab inverse of L → Pn , aij 7→ Aij.
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